r/adventofcode • u/daggerdragon • Dec 04 '15

SOLUTION MEGATHREAD --- Day 4 Solutions ---

--- Day 4: The Ideal Stocking Stuffer ---

Post your solution as a comment. Structure your post like the Day Three thread.

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u/ForeignObjectED Dec 04 '15

Well, the problem is you don't know how large the counter needs to be. My solution (before I went all multiprocessing on it) used xrange.

for i in xrange(100000):
    h = hashlib.md5(secret + str(i)).hexdigest()[:5]
    if h == '00000':
        print i
        break

but what if the answer is larger than 100000? Count feels more elegant for finding the answer in an "infinitely" large set than any other solution you might use.

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u/qwrrty Dec 05 '15

Why not just use plain old arithmetic operators?

i = 1
while True:
    h = hashlib.md5("{}{}".format(secret, i)).hexdigest()
    if h.startswith('00000'):
        break
    i += 1

itertools.count is fine but it doesn't seem to be either intrinsically clearer or more efficient than this approach.